On representations of Heisenberg groups Let $G$ be the group of upper triangular $3 \times 3$ matrices with 1's in the diagonal and entries in $\mathbb{F}_3$. I want to find and irreducible complex representation of dimension 3. 
What would be an easy way to see this?
Now if instead of asking that the entries belong to $\mathbb{F}_3$ we ask that they belong to $\mathbb{Z}/n\mathbb{Z}$, is there a complex irreducible representation of dimension $n$? In that case, what would be such a representation?
 A: Let $G$ be your group and let $H$ be the subgroup of matrices of the form $\begin{pmatrix}1&0&*\\0&1&*\\0&0&1\end{pmatrix}$, and let $Z$ be the subgrouo of matrices of the form $\begin{pmatrix}1&0&*\\0&1&0\\0&0&1\end{pmatrix}$. 
Pick any character $\chi:H\to\mathbb C^\times$ whose restriction to $Z$ is nontrivial. For example, we can pick $\chi$ so that $\chi\left(\left(\begin{smallmatrix}1&0&a\\0&1&b\\0&0&1\end{smallmatrix}\right)\right)=\omega^a$, with $\omega$ a primitive cube root of $1$ in $\mathbb C$.
Consider the induced representation $V=\chi\uparrow^G_H$. This has dimension $3$ because the index of $H$ in $G$ is $3$. The restriction $V|_Z$ is a direct sum of three copies of $\chi$. As $Z$ acts trivially on any $1$-dimensional representation of $G$ (because $Z$ is the derived subgroup of $G$) this tells us that $V$ does not have any $1$-dimensional $G$-submodules. As $G$ has only $1$-dimensional submodules ands $3$-dimensional ones, $V$ must be a $3$-dim irrep of $G$.

If one works this out explicitly (and if I did not make any mistake!) we can describe this module as follows. Pick a primitive cube root of unity $\omega$ and consider the vector space with basis $\{e_x:x\in\mathbb F_3\}$ indexed by the elements of $\mathbb F_3$. The action of a matrix in the group $G$ on $V$ is then such that: $$\left(\begin{smallmatrix}1&a&b\\0&1&c\\0&0&1\end{smallmatrix}\right)\cdot e_x=\omega^{b-(x+a)c}e_{x+a}.$$ One has two options for $\omega$ and in this way we obtain the two $3$-dimensional irreps.
